F.5 M2 __integration

∫ sin3 x dx/(1 + cos4 x) = -∫ sin2 x d(cos x)/(1 + cos4 x)

= -∫ (1 - cos2 x) d(cos x)/(1 + cos4 x)

= -∫ (1 - u2) du/(1 + u4) where u = cos x

= ∫ (u2 - 1) du/(u4 + 1)

For (u2 - 1)/(u4 + 1), by u4 + 1 = (u2 + 1)2 - 2u2 = (u2 + √2u + 1)(u2 - √2u + 1), we can divide it into partial fractions that:

(u2 - 1)/(u4 + 1) = (Au + B)/(u2 + √2u + 1) + (Cu + D)/(u2 - √2u + 1) where A, B, C and D are constants.

Then:

u2 - 1 = (Au + B)(u2 - √2u + 1) + (Cu + D)(u2 + √2u + 1)

= (A + C)u3 + (B + D - √2A + √2C)u2 + (A + C - √2B + √2D)u + (B + D)

Comparing the coeff, we have:

A + C = 0

B + D - √2A + √2C = 1

A + C - √2B + √2D = 0

B + D = -1

Solving, we have A = -1/√2, B = -1/2, C = 1/√2 and D = -1/2 giving: ∫ (u2 - 1) du/(u4 + 1) = - (1/2) ∫ (√2u + 1)du/(u2 + √2u + 1) + (1/2) ∫ (√2u - 1)du/(u2 - √2u + 1) = - (√2/4) ∫ (2u + √2)du/(u2 + √2u + 1) + (√2/4) ∫ (2u - √2)du/(u2 - √2u + 1)= - (√2/4) ∫ d(u2 + √2u + 1)/(u2 + √2u + 1) + (√2/4) ∫ d(u2 - √2u + 1)/(u2 - √2u + 1)= - (√2/4) ln (u2 + √2u + 1) + (√2/4) ln (u2 - √2u + 1) + C= - (√2/4) ln (cos2 x + √2 cos x + 1) + (√2/4) ln (cos2 x - √2 cos x + 1) + C

嗯嗯 我也做出這答案
只是太複雜了
一度懷疑題目出錯了><"